Engineering Mathematics Test 12

Explanation:

\(f\left( {{x_n}} \right) = x_n^3 - 23\)

\(f'\left( {{x_n}} \right) = 3x_n^2\)

According to Newton-Raphson formula,

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

\({x_{n + 1}} = {x_n} - \left( {\frac{{x_n^3 - 23}}{{3x_n^2}}} \right)\)

Concept:

Newton Raphson Method:

\(\begin{array}{l} {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\ \end{array}\)  

Calculation:

Given:

f(x) = x² – 3x + 3

f (x₀ = 0) = 3

f’(x) = 4x – 3

f’ (x₀ = 0) = - 3

By Newton Raphson method

\({x_i} = {x_o} - \frac{{f\left( {{x_o}} \right)}}{{f'\left( {{x_o}} \right)}}\)

\({x_1} = 0 - \frac{3}{{ - 3}}\)

x₁ = 1

\(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx\)

number of intervals = 4

\(h = \frac{1}{4} = 0.25\)

By Trapezoidal rule,

\(\mathop \smallint \limits_{{x_0}}^{{x_0} + nh} f\left( x \right)\;dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + \ldots + {y_{n - 1}}} \right)} \right]\)

\(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\)

\(= \frac{1}{8}\left[ {\left( {1 + 0.5} \right) + 2\left( {0.8 + 0.66 + 0.57} \right)} \right]\)

= 0.695

Explanation:

Let x₀ = 0, x₁ = 2

f(x₀) = -3 = -ve, f(x₁) = 1 = +ve

So, root lies between 0 and 2

1^st iteration:

\({x_2} = \frac{{0 + 2}}{2} = 1\)

f(x₂) = 1 – 3 = -2 (-ve)

So, root lies between 1 and 2

2^nd iteration:

\({x_3} = \frac{{1 + 3}}{2}\)

∴ x₃ = 1.5

Explanation:

Given:

\(\frac{{dy}}{{dx}} - \left( {xy + {x^2}} \right) = 0\)

\(\frac{{dy}}{{dx}} = xy + {x^2} = f\left( {x,\;y} \right)\)

Concept:

Euler’s first order formula is

y_n+1 = y_n + h ⋅ f(x_n, y_n)

y₁ = y₀ + h ⋅ f (x₀, y₀)       ----(1)

f(x₀, y₀) = 0 + 0 = 0

h = 1

y₀ = 0

y₁ = 0 + 1 ⋅ (0) = 0

x₁ = x₀ + h = 0 + 1 = 1

f(x₁, y₁) = 1(0) + 1² = 1

y₂ = y₁ + h ⋅ f(x₁, y₁)

y₂ = 0 + 1 ⋅ (1) = 1

x₂ = x₁ + h = 1 + 1 = 2

Now,

y₃ = y₂ + h ⋅ f(x₂, y₂)

f(x₂, y₂) = 2(1) + 2² = 6

∴ y₃ = 1 + 1.(6) = 7

Explanation:

Equation are

20x + y – 2z = 17

3x + 20y – z = -18

2x – 3y + 20z = 25

Re-writing the above equations.

\(x = \frac{1}{{20}}\left( {17 - y + 2z} \right)\)         ---(1)

\(y = \frac{1}{{20}}\left( { - 18 - 3x + z} \right)\)        ---(2)

\(z = \frac{1}{{20}}\left( {25 - 2x + 3y} \right)\)          ---(3)

Putting y₀ = z₀ = 0 in equation (1)

\({x_1} = \frac{1}{{20}}\left( {17 - 0 + 0} \right)\)

∴ x₁ = 0.85

Putting x = 0.85, z = 0 in equation (2)

\({y_1} = \frac{1}{{20}}\left( { - 18 - \left( {3 \times 0.85} \right) + 0} \right)\)

∴ y₁ = - 1.0275

Putting x = 0.85, y = -1.0275 in eq. (3)

\({z_1} = \frac{1}{{20}}\left[ {25 - \left( {2 \times 0.85} \right) + 3\left( { - 1.0275} \right)} \right]\)

∴ z₁ = 1.010875

Concept:

According to Simpson’s (1/3)rd rule,

The estimated volume of the reservoir

\( = \frac{1}{3} \times h\left\{ {{A_0} + {A_6} + 4\left( {{A_1} + {A_3} + {A_5}} \right) + 2\left( {{A_2} + {A_4}} \right)} \right\}\)

where

h = interval at which cross – section are taken

A₀, A₁, A₂, ……. A₆ are the seven cross-sections taken at 9 m interval.

Calculations:

The volume of the reservoir (V)

\(V = \frac{1}{3} \times 9\left[ {210 + 170 + 4\left( {250 + 350 + 230} \right) + 2\left( {320 + 290} \right)} \right]\)

\(V = \frac{1}{3} \times 9 \times 4920\)

V = 14760 m³

Explanation:

f(x) = x sin x – 1

f(1) = -0.158529 < 0

f(1.5) = 0.496242 > 0

So root will lie between 1 & 1.5

\({x_1} = \frac{1}{2}\left( {1 + 1.5} \right) = 1.25\)

f(x₁) = 0.18623 > 0

so root lies between 1 & 1.25

\({x_2} = \frac{1}{2}\left( {1 + 1.25} \right) = 1.125\)

f(x₂ ) = 0.01505 > 0

So root lies between 1 & 1.125

\({x_3} = \frac{{1 + 1.25}}{2} = 1.0625\)

f(x₃) = -0.0718 < 0

f(x₂) > 0 & f(x₃) < 0

So root lies between x₂ & x₃

Concept:

\(\frac{{dy}}{{dx}} = f\left( {x,y} \right),\;y\left( {{x_0}} \right) = {y_0}\)

Runge-Kutta Method of first order:

k₁ = h f(x₀, y₀)

y₁ = y₀ + k where k = k₁ = h f (x₀, y₀)

y₂ = y₁ + k where k = h f (x₁, y₁)

Runge-Kutta Method of Second order:

k₁ = h f(x₀, y₀)

k₂ = h f (x₀ + h, y₀ + k₁)

\(k = \frac{{{k_1} + {k_2}}}{2}\)

y₁ = y₀ + k

Runge-Kutta Method of Third order:

k₁ = h f (x₀, y₀)

\(\begin{array}{l} {k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right)\\ k' = hf\left( {{x_0} + h,{y_0} + {k_1}} \right) \end{array}\)  

\(\begin{array}{l} {k_3} = hf\left( {{x_0} + h,{y_0} + k'} \right)\\ k = \frac{1}{6}\left( {{k_1} + 4{k_2} + {k_3}} \right) \end{array}\)  

y₁ = y₀ + k

Runge-Kutta Method of fourth order:

k₁ = h f (x₀, y₀)

\(\begin{array}{l} {k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right)\\ {k_3} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_2}}}{2}} \right) \end{array}\)  

\(\begin{array}{l} {k_4} = hf\left( {{x_0} + h,{y_0} + {k_3}} \right)\\ k = \frac{1}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right) \end{array}\)  

y₁ = y₀ + k

Calculation:

Question is about third order Runge Kutta method with step-size h = 0.1.

\(\frac{{dy}}{{dx}} = - y = f\left( {x,y} \right);{x_0} = 0,\;{y_0} = 1\)

f(x₀, y₀ ) = f(0, 1) = -1

k₁ = h f (x₀, y₀) = 0.1 × (-1) = -0.1

\({k_2} = hf\left( {{x_0} + \frac{h}{2},{y_0} + \frac{{{k_1}}}{2}} \right) = hf\left( {0.05,\;0.95} \right)\)

= 0.1 [-0.95] = -0.095

k’ = h f (x₀ + h, y₀ + k₁) = h f (0.1, 0.9) = =0.9 h

k’ = -0.09

k₃ = h f (x₀ + h, y₀ + k’) = h f (0.1, 0.91) = -0.91 h

k₃ = -0.091

\(\begin{array}{l} k = \frac{1}{6}\left( {{k_1} + 4{k_2} + {k_3}} \right)\\ = \frac{{ - 1}}{6}\left( {0.1 + 4\left( {0.095} \right) + 0.091} \right) = - 0.095167 \end{array}\)  

Explanation:

Let f(x) = x⁴ – x – 10

f'(x) = 4x³ – 1

Newton-Raphson’s formula is

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Putting n = 0, first approximation is given by

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)     ---(1

Since, x₀ = 2

f(x₀) = 2⁴ – 2 – 10 = 4

f’(x₀) = 31

Put in (1)

\({x_1} = 2 - \frac{4}{{31}}\)

∴ x₁ = 1.8709

Also,

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\)     ---(2)

f(x₁) = (1.8709)⁴ – 1.8709 – 10 = 0.3809

f’(x₁) = 4(1.8709)³ – 1 = 25.1945

\({x_2} = 1.8709 - \frac{{0.3809}}{{25.1945}} = 1.8557\)